What is the extreme f(x,y)=x^2+y^2-6y+9 ?

The extreme f(x,y)=x^2+y^2-6y+9 is Minimum(0,3)

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extreme f(x,y)=x^2+y^2-6y+9

extreme points

f (x, y) = x^{2} + y^{2} - 6 y + 9

Minimum (0, 3)

Solution steps

Find the critical points:

(0, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 3) :

Minimum

Minimum (0, 3)

f (x) = x^{2}, - 1 \leq x \leq 0

3 x^{2} - 5 x + 2

f (x) = x^{4} - 4 x^{3} + 16 x + 4

f (x) = x^{7} - 7 x

f (x, y) = 18 x^{2} - 32 y^{2} - 36 x - 128 y - 110

What is the extreme f(x,y)=x^2+y^2-6y+9 ?
The extreme f(x,y)=x^2+y^2-6y+9 is Minimum(0,3)